Problems on Distance Between Two Points



Solving the problems on distance between two points with the help of the formula, in the below examples use the formula to find distance between two points.


Worked-out problems on distance between two points:

1. Show that the points (3, 0), (6, 4) and (- 1, 3 ) are the vertices of a right-angled Isosceles triangle. 

Solution:
 Let the given points be A(3, 0), B (6, 4) and C (-1, 3). Then we have, 

AB² = (6 - 3)² + (4 - 0)² = 9 + 16 = 25; 

BC² = (-1 - 6)² + (3 - 4 )² = 49 + 1= 50 

and CA² = (3 + 1)² + (0 - 3)² = 16 + 9= 25. 

From the above results we get, 

AB² = CA² i.e., AB = CA

which proves that the triangle ABC is isosceles. 

Again, AB² + AC² = 25 + 25 = 50 = BC² 

which shows that the triangle ABC is right-angled. 

Therefore, the triangle formed by joining the given points is a right-angled isosceles triangle. Proved



2. If the three points (a, b), (a + k cos α, b + k sin α) and (a + k cos β, b + k sin β) are the vertices of an equilateral triangle, then which of the following is true and why ?

(i) | α - β| = π/4

(ii) |α - β| = π/2

(iii) |α - β| = π/6

(iv) |α - β| = π/3

Solution:
Let the vertices of the triangle be A (a, b), B (a + k cos α, b + k sin α) and C (a + k cos β, b + k sin β).

Now, AB² = (a + k cos α - a)² + (b + k sin α - b)²

= k² cos² α + k² sin² α = k²;

Similarly, CA² = k² and

BC² = (a + k cos β - a - k cos α)² + (b + k sin β - b - k sin α)²

= k² (cos² β + cos² α - 2 cos α cos β + sin² β + sin² α - 2 sin α sin β)

= k² [cos² β + sin² β + cos² α + sin² α - 2(cos α cos β + sin α sin β)]

= k² [1 + 1 - 2 cos (α - β)]

= 2k² [1 - cos (α - β)]

Since ABC is an equilateral triangle, hence

AB² = BC²

or, k² = 2k² [1 - cos (α - β)]

or, 1/2 = 1 - cos(α - β) [since, k # 0]

or, cos (α - β) = 1/2 = cos π/3

Therefore, |α - β| = π/3 .

There for, condition (iv) is true.



3. Find the point on the y-axis which is equidistant from the points (2, 3)and(-1, 2).

Solution:
Let P(0, y) be the required point on the y-axis and the given points are A (2, 3) and B(- 1, 2). By question,

PA = PB = PA² = PB²

or, (2 - 0)² + (3 - y)² = (-1 - 0)² + (2 – y)²

or, 4 + 9 + y² - 6y = 1 + 4 + y² - 4y

or, - 6y + 4y = 1 - 9 or, - 2y = -8

or, y = 4.

Therefore, the required point on the y-axis is (0, 4).



4. Find the circum-centre and circum-radius of the triangle whose vertices are (3, 4), (3, - 6) and (- 1, 2). 


Solution:
 Let A(3, 4), B (3, - 6), C (- 1, 2) be the vertices of the triangle and P(x, y ) the required circum-centre and r the circum-radius. Then, we must have, 

r² = PA² = (x - 3)² + (y - 4)² ……………………..(1) 

r² = PB² = (x - 3)² + (y + 6)² ……………………….(2) 

and r² = PC² = (x + 1)² + (y - 2)² ……………………….(3) 

From (1) and (2) we get, 

(x - 3)² + (y - 4)² = (x - 3)² + (y + 6)² 

Or, y² - 8y + 16 = y² + 12y + 36 

or, - 20y = 20 or, y = - 1 

Again, from (2) and (3) we get, 

(x - 3)² + (y + 6)² = (x + 1 )² + (y - 2)²

or, x² - 6x + 9 + 25 = x² + 2x + 1 + 9 [putting y = - 1] 

or, - 8x = - 24 

or, x = 3 

Finally, putting x = 3 and y = - 1 in (1) we get, 

r² = 0² + (-1 - 4)² = 25 

Therefore, r = 5 

Therefore, the co-ordinates of circum-centre are (3, - 1) and circum-radius = 5 units. 



5. Show that the four points (2, 5), (5, 9), (9, 12) and (6, 8) when joined in order, form a rhombus. 

Solution:
 Let the given points be A(2, 5), B (5, 9), C (9, 12) and D(6, 8). Now, AB² = (5 - 2)² + (9 - 5)² = 9 + 16 = 25

BC² = (9 - 5)² + (12 - 9)² = 16 + 9 = 25

CD² = (6 - 9)² (8 - 12)² = 9 + 16 = 25

DA² = (2 - 6)² + (5 - 8)² = 16 + 9 = 25

AC² = ( 9 - 2)² + (12 - 5)² = 49 + 49 = 98

and BD² = (6 - 5)² + (8 - 9)² = 1 + 1 = 2

From the above result we see that

AB = BC = CD = DA and AC ≠ BD

That is the four sides of the quadrilateral ABCD are equal but diagonals AC and BD are not equal. Therefore, the quadrilateral ABCD is a rhombus. Proved.

The above worked-out problems on distance between two points are explained step-by-step with the help of the formula.



 Co-ordinate Geometry 




11 and 12 Grade Math 

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